Combustion of $1.78 g$ of an organic compound (C, H, N, O only) gives $2.64 g$ $CO X_{2}$ and $1.26 g$ $H X_{2} O$ . In a Kjeldahl analysis, all the nitrogen in $0.890 g$ of the compound is converted to $NH X_{3}$ , which is neutralised by $25.0 mL$ of $0.200 M$ $H X_{2} SO X_{4}$ . The empirical formula is (C = 12, H = 1, N = 14, O = 16)

A $C X_{3} H X_{7} NO$
B $C X_{3} H X_{7} NO X_{2}$
C $C X_{3} H X_{5} NO X_{2}$
D $C X_{2} H X_{5} NO X_{2}$

Answer: B

In 1.78 g: C = 2.64/44 = 0.06 mol; H = 2 x 1.26/18 = 0.14 mol. Kjeldahl: n(H2SO4) = 0.025 x 0.200 = 0.00500 mol, so n(NH3) = 0.0100 mol N in 0.890 g, hence 0.0200 mol N in 1.78 g. Mass of O = 1.78 - 0.72 - 0.14 - 0.28 = 0.64 g = 0.04 mol. Ratio C : H : N : O = 3 : 7 : 1 : 2.

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Answer: B

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